Show AH Work!!! 1. Find the probability value: (15 points = 5 points each) a) P(-0.32 5 z 51.62)= b) P(z>-1.03)= e) P(zg-1.35)= 2. Find the value of 20. such that: a) P( -z:: E z 5 20) = 0.82 (5 points) b) PG: E 20) = 0.9082 (5 points) 3. The random variable X is normally distributed with mean 80 and standard deviation 12. a) "What is the probability that a 1Jalue of X chosen at random will be between 7'0 and 85'? (10 points) b) What is the probability that a value of X chosen at random will be less than 75? (IllI points) 4. The scores on a placement test have a mound-shaped distribution with mean 400 and standard deviation 45. a) What percentage of people taking this exam will have scores of 390 or greater? (10 points) b) What percentage of the people taking this test will have scores between 415 and 435'? (10 points) c) Mary is told that 90.82% of the students taking the test have higher scores than she does? What was Mary's score? (10 points) 5. A population of 40 year-old females has a mean salary of $3 9.321 with a standard deviation of $2.120. If a sample of 100 women is taken: a) What is the probability their mean salaries will be less than $38,000? (10 points) b) What is the probability their mean salaries will be between $38,500 and $39,000? (10 points) 6. Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). Find the probability that a randomly selected adult has an IQ between 95 and 110 (somewhere in the range of normal to bright normal). (5 points) Formulas Discrete Probability Distribution Mean E(X ) = u = >xP(x) all x Variance v(X)=o' = >(x -u)'P(x) Standard Deviation 0 = VO2 all x Binomial Distribution If X ~ Bin(n, p), then P(X = x) -(20*(1-pax-. n! x!(n-x)[ P (1-p)"-* x=0,1,...,n. Computing Standard Scores Individual Z = - data value-mean standard deviation Group Z = data value-mean standard deviation/ Vn Hypothesis Testing Mean Z = x - H Proportion Z =- P-P pq V In